Problem: Find the sum of $46 + 42 + 38 +... + (-446) + (-450)$.
Explanation: Getting started We're dealing with an arithmetic series because the difference between terms is constant. That is, each term is $4$ less than the one before it. We need a formula to compute the sum of the terms. Formula for arithmetic series The sum $S_n$ of a finite arithmetic series is $S_n = \dfrac {\left(a_1 + a_n \right)}{2} \cdot n$ where $a_1$ is the first term, $a_n$ is the last term, and $n$ is the number of terms. What do we need to use the formula? The first term $(a_1 = {46})$ and the last term $(a_n = {-450})$ are given in the question. We need to find $n$ (the number of terms). Step 1: Find $n$ (the number of terms) The sequence decreases by $46 - (-450) = 496$ from the first term to the last term. Because the sequence decreases by $4$ each time, it takes $\dfrac{496}{4} = 124$ terms to get from the first term to the last term. We still need to count the first term, so there are $124 + 1 = {125}$ terms in the sequence. In other words, $n = {125}$. Step 2: Find the sum $(S_n)$ of the series $\begin{aligned} S_n &= \dfrac {\left(a_1 + a_n \right)}{2} \cdot n \\\\ S_{{125}}&= \dfrac {\left({46} + ({-450}) \right)}{2} \cdot {125} \\\\ S_{{125}} &= -202 \left(125\right) \\\\ S_{{125}} &= -25{,}250\end{aligned}$ The answer $ -25{,}250 $